CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-23 (gpt-6.1-sol)

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Conservative momentum alone does not fix vacuum velocity

Physical scope

Formal weak mass and momentum laws do not determine velocity on vacuum; this is not a claim that a material fluid occupies vacuum.

Statement refuted

Assume Countable Choice: every sequence of nonempty sets admits a choice function. Use the restricted Euclidean Lebesgue spaces and almost-everywhere convention of the conservative weak-form definition. On a fixed nonempty bounded open spatial domain $Ω \subseteq R^{3}$ and a finite time interval $(0, T_{f})$ with $T_{f} > 0$ , the weak conservative mass and momentum equations, with the density, stress and body-acceleration fields held fixed and with fixed initial conservative mass and momentum data, determine at most one bounded velocity field up to almost-everywhere equality, even where the density is zero.

Statement references

Conservative weak equations and initial conservative data

Facts & Assumptions

Given: The fixed open domain, finite time interval, zero density/stress/body source, zero initial conservative data, and the two bounded velocity fields in the Counterexample.

[A1]

Use the mathematical Countable Choice and Lebesgue-measure convention of Locally integrable conservative weak and initial forms.

[F1]

Locally integrable conservative weak and initial forms: for $U, F, S \in L_{l o c}^{1}$ , the weak equation is the compactly supported $C^{1}$ test identity with the initial conservative datum; the mass and momentum instances use $(U, F, S) = (ρ, ρ u, 0)$ and $(ρ u, ρ u \otimes u - σ, ρ b)$ .

Counterexample

Let $Ω = (0, 1)^{3}$ metres and $I = (0, T_{f})$ with $T_{f} > 0$ . The weak conservative mass and momentum equations use mass density $ρ$ , velocity $u$ , momentum density $ρ u$ , mass flux $ρ u$ , momentum flux $ρ u \otimes u - σ$ , and momentum source $ρ b$ , tested against compactly supported $C^{1}$ scalar and vector fields with the conservative initial-data terms. Set $ρ \equiv 0$ , $σ \equiv 0$ , $b \equiv 0$ , and initial mass and momentum densities equal to zero. Both the velocity field $u_{1} \equiv 0$ and any fixed nonzero constant velocity field $u_{2} \equiv v_{*}$ (with $v_{*} \in R^{3}$ in metres per second) are bounded and smooth. The two choices have the same zero conservative mass and momentum variables and satisfy the same weak mass and momentum equations, although $u_{1} \neq u_{2}$ everywhere. Thus those conservative equations alone do not determine a velocity on the set where $ρ = 0$ . The example concerns the formal vacuum degeneracy of the equations, not a fluid-particle velocity in physical vacuum, and makes no claim about an energy or constitutive equation.

1.1givenalgebra

For both $u_{1}$ and $u_{2}$ , $ρ = 0$ , so the mass density, mass flux, momentum density, quadratic momentum flux, momentum source and initial conservative data are all identically zero. The fields and all products are locally integrable on the finite space-time domain.

2.1givenA1F1step 1.1

In each weak test identity from [F1], every space-time integrand and every initial term is therefore zero for either velocity. Both choices satisfy the same weak mass and momentum equations.

3.1givenstep 1.1step 2.1∎

The constant fields $u_{1}$ and $u_{2}$ differ while their conservative mass and momentum data and fluxes coincide. Hence the weak mass and momentum equations alone do not determine $u$ on ${ρ = 0}$ . No division by density or assertion about a material velocity in vacuum is made.

Conservative momentum alone does not fix vacuum velocity

Statement refuted

Statement references

Facts & Assumptions

Counterexample

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